Question

If \(z=x+iy,\; z^{1/3}=a-ib\), then \(\dfrac{x}{a}-\dfrac{y}{b}=k(a^2-b^2)\), where \(k\) is equal to:

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Always expand powers of complex numbers using binomial theorem.

Answer 1

The Correct Option is C

Step 1: Given \(z^{1/3}=a-ib\Rightarrow z=(a-ib)^3\).
Step 2: \[ z=(a^3-3ab^2)-i(3a^2b-b^3) \] So, \[ x=a^3-3ab^2,\quad y=3a^2b-b^3 \]
Step 3: \[ \frac{x}{a}-\frac{y}{b}=a^2-3b^2-(3a^2-b^2)=3(a^2-b^2) \] Thus \(k=3\).